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Question Number 110288 by mathdave last updated on 28/Aug/20
Answered by bemath last updated on 28/Aug/20
Δ((be)/(math))▽ let y = vx ⇒dy=v dx+ x dv ⇔(4vx−2x)dx=(x+vx)(v dx+x dv) (4v−2)dx=(1+v)(v dx + x dv ) (4v−2)dx=(v+v^2 )dx + (x+vx)dv (3v−2−v^2 ) dx = x(1+v)dv ⇔ −(dx/x) = (((1+v)dv)/(v^2 −3v+2)) ⇔−ln x+ c = ∫ (((1+v)dv)/((v−1)(v−2))) ⇔−ln x+c = 3ln (v−2)−2ln (v−1) ln ((C/x))= ln ((((v−2)^3 )/((v−1)^2 ))) ⇔ C(v−1)^2 =x(v−2)^3 C(((y−x)/x))^2 =x(((y−2x)/x))^3 C(y−x)^2 =(y−2x)^3 y(1)=3⇔ C(3−1)^2 =(3−2)^3 4C = 1 →C=(1/4) ∴ (y−x)^2 =4(y−2x)^3
$$\:\:\:\:\:\Delta\frac{{be}}{{math}}\bigtriangledown \\ $$$${let}\:{y}\:=\:{vx}\:\Rightarrow{dy}={v}\:{dx}+\:{x}\:{dv} \\ $$$$\Leftrightarrow\left(\mathrm{4}{vx}−\mathrm{2}{x}\right){dx}=\left({x}+{vx}\right)\left({v}\:{dx}+{x}\:{dv}\right) \\ $$$$\left(\mathrm{4}{v}−\mathrm{2}\right){dx}=\left(\mathrm{1}+{v}\right)\left({v}\:{dx}\:+\:{x}\:{dv}\:\right) \\ $$$$\left(\mathrm{4}{v}−\mathrm{2}\right){dx}=\left({v}+{v}^{\mathrm{2}} \right){dx}\:+\:\left({x}+{vx}\right){dv} \\ $$$$\left(\mathrm{3}{v}−\mathrm{2}−{v}^{\mathrm{2}} \right)\:{dx}\:=\:{x}\left(\mathrm{1}+{v}\right){dv} \\ $$$$\Leftrightarrow\:−\frac{{dx}}{{x}}\:=\:\frac{\left(\mathrm{1}+{v}\right){dv}}{{v}^{\mathrm{2}} −\mathrm{3}{v}+\mathrm{2}} \\ $$$$\Leftrightarrow−\mathrm{ln}\:{x}+\:{c}\:=\:\int\:\frac{\left(\mathrm{1}+{v}\right){dv}}{\left({v}−\mathrm{1}\right)\left({v}−\mathrm{2}\right)} \\ $$$$\Leftrightarrow−\mathrm{ln}\:{x}+{c}\:=\:\mathrm{3ln}\:\left({v}−\mathrm{2}\right)−\mathrm{2ln}\:\left({v}−\mathrm{1}\right) \\ $$$$\mathrm{ln}\:\left(\frac{{C}}{{x}}\right)=\:\mathrm{ln}\:\left(\frac{\left({v}−\mathrm{2}\right)^{\mathrm{3}} }{\left({v}−\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\Leftrightarrow\:{C}\left({v}−\mathrm{1}\right)^{\mathrm{2}} ={x}\left({v}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$${C}\left(\frac{{y}−{x}}{{x}}\right)^{\mathrm{2}} ={x}\left(\frac{{y}−\mathrm{2}{x}}{{x}}\right)^{\mathrm{3}} \\ $$$${C}\left({y}−{x}\right)^{\mathrm{2}} =\left({y}−\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$$${y}\left(\mathrm{1}\right)=\mathrm{3}\Leftrightarrow\:{C}\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{3}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\mathrm{4}{C}\:=\:\mathrm{1}\:\rightarrow{C}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\therefore\:\left({y}−{x}\right)^{\mathrm{2}} =\mathrm{4}\left({y}−\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$
Commented by mathdave last updated on 28/Aug/20
good work
$${good}\:{work} \\ $$

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